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How to calculate the resistance when the watt of motor is known
Without considering the power factor
P=UI COSα=1
R = u / I, a problem of Ohm's law!
If the bulbs L1 and L2 marked with "6V 6W" are connected to the 6V power supply, the lamp with higher actual power is the one with higher power______ (select "L1" and "L2"), the current of the main circuit in the circuit is______ A．
(1) The two lamps are connected in parallel at both ends of the 6V power supply, the actual voltage is equal to the rated voltage, so the actual power is equal to the rated power, so the bulb L1 is brighter; (2) the current through the two lamps is I1 = p1u = 6w6v = 1A; I2 = P2u = 3w6v = 0.5A, so the trunk current is I = I1 + I2 = 1A + 0.5A = 1.5A, so the answer is: L1; 1.5
Connect two small bulbs marked "6V 3W" and "6V 2W" in series and connect them to the two ends of the 6V power supply. What is the total power of the two small bulbs?
Hurry!
Total resistance of two small bulbs in series:
According to: r = u × U / P
R total = 6 × 6 / 3 + 6 × 6 / 2 = 30 (Ω)
When connected to a 6V power supply, the total power of two small bulbs in series is as follows:
P total = u × U / r = 6 × 6 / 30 = 1.2 (W)
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